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jbritto

  • 2 years ago

What is the limit of (1+cosx)/sinx when x approaches 0

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  1. zugzwang
    • 2 years ago
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    l'Hôpital's rule, are you allowed to use that?

  2. zugzwang
    • 2 years ago
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    Wait, I panicked... no l'Hôpital's rule, I'm sorry

  3. myininaya
    • 2 years ago
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    How can you use l-hospital here?

  4. myininaya
    • 2 years ago
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    You have 2/0 not 0/0

  5. myininaya
    • 2 years ago
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    \[\lim_{x \rightarrow 0}\frac{1+\cos(x)}{\sin(x)}\] Recall the following: \[\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1 \text{ , } \lim_{x \rightarrow 0}\frac{\cos(x)-1}{x}=0\] Let's see if we can use that here: \[\lim_{x \rightarrow 0}\frac{\cos(x)+1}{\sin(x)} \cdot \frac{\cos(x)-1}{\cos(x)-1}\] \[\lim_{x \rightarrow 0} \frac{\cos^2(x)-1}{\sin(x)(\cos(x)-1)}\] \[\lim_{x \rightarrow 0}=\frac{-\sin^2(x)}{\sin(x)(\cos(x)-1)}\] \[\lim_{x \rightarrow 0}\frac{-\sin(x)}{\cos(x)-1}\] Now try it :)

  6. myininaya
    • 2 years ago
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    Using L'hospital you will get something easier to look at :)

  7. myininaya
    • 2 years ago
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    We didn't actually need any of those things I told you to recall above :)

  8. jbritto
    • 2 years ago
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    Thanks. It still becomes -1/0 which is undefined ??

  9. shubhamsrg
    • 2 years ago
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    its not -1/0 but its 0/0 @jbritto sin0 = 0 cos0 =1

  10. shubhamsrg
    • 2 years ago
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    anyways,,there is an alternative also.. 1+ cosx = 1 + 2 cos^2 (x/2) -1 = 2cos^2 (x/2) and sinx = 2 sin(x/2) cos(x/2) so whole expression is 2 cos^2(x/2) / 2 sin(x/2) cos(x/2) = cos(x/2)/ sin(x/2) = cot (x/2) not plug in x=0

  11. shubhamsrg
    • 2 years ago
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    oh,,so you were saying it becomes 1/0 !! ohh,,am sorry.. in that case either the limit is undefined or you maybe made a typing error//

  12. jbritto
    • 2 years ago
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    yes. applying L-Hosp rule, it will become -cosx/sinx which is -1/0 undefined

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